Question

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24. A better model for two indistinguishable fair dice? You may be uncomfortable with the sample space given in class for the possible outcomes of the roll of a pair of fair dice because when dealing with two dice you cannot, or at least you do not, distinguish between them. Thus for example, the two outcomes (1,3) and (3,1) should perhaps be considered as the same outcome. This argument leads to the following 21-member sample space: (23>)A56.6 (2,2) (2,3) (2,4) (2,5) (2,6) (3,3) (3,4) (3,5) (3,6) (4,4) (4,5) (4,6) (5,5) (5,6) (6,6) a) Using this sample space and assuming that all 21 outcomes are equally likely, compute the following probabilities i) Both dice show numbers greater than 3. ii) The sum of the two numbers is 6 iii) The two dice show the same number iv) At least one 4 shows. v) At least one number greater than 4 shows b) Why do you think the model presented in class is the one that is successfully used in practice and this one is not? (It must be that it gives more accurate predictions of long-run relative frequencies; why do you think that this is true?) c) How could you convince someone (perhaps yourself?) empirically that the class model is better by using the results of, say, 25,200 rolls of two fair dice? (Note 25,200 is a multiple of both 21 and 36.) d) It turns out that the 21-member sample space is not useless for the roll of two fair dice, however, if we assign the appropriate probabilities to its members (not 1/21 each), it works perfectly well. What are the appropriate probabilities? (Be certain that your 21 probabilities add

Answer 1

$(a)~i.\\ Probability~that~both~dice~show~numbers~greater~than~3\\ =P(\{(4,5),(4,6),(5,6)\})=\frac{3}{21}=1/7.\\ ii.~Probability~that~the~sum~of~the~two~numbers~is~6\\ =P(\{(1,5),(2,4),(3,3)\})=\frac{3}{21}=1/7.\\ iii.~Probability~that~two~dice~show~the~same~number\\ =p(\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\})=6/21=3/7\\ iv.~Probability~that~at~least~4~shows\\ =P(\{(1,4),(2,4),(3,4),(4,4),(4,5),(4,6)\})=6/21=3/7\\$

$v.~Probability~that~at~least~one~number~greater~than~4~shows\\ =P(\{(1,5),(1,6),(2,5),(2,6),(3,5),(3,6),(4,5),(4,6),(5,5),(5,6),(6,6)\})=11/21.$

$(b)~Since~we~ignore~(i,j),~i>j~i.e.~(2,1),(3,2)...etc~so~we~expect~that~\\ P(\{(i,j)\})=2P(\{(i,i)\})~as~i<j~but~here~P(\{(i,j)\})=P(\{(i,i)\})~\forall~i,j.$

$(d)~We~attach~P(\{(i,j)\})=\frac{1}{18}~i<j=1(1)6\\ P(\{(i,i)\})=1/36,~i=1,2,3,4,5,6\\ then~\sum_{i=1}^6P(\{(i,i)\})&plus;\sum_{i<j=1}^6P(\{(i,j)\})=\frac{\binom{6}{1}}{36}&plus;\frac{\binom{6}{2}}{18}=1\\$

$(c)~We~expect~that~P(\{i,j\})~occurs~equal~number~of~times~when~we~repeat\\ this~process~large~no.~of~time.~However~if~we~consider~only~21~sample~points\\ then~each~(\{i,i\})~occurs~1200~times~and~each~(\{i,j\}),~i<j~also~occurs\\ 1200~times.~However~if~the~die~is~fair~then~it~is~expected~that~the~outcome~of~\\ (i,j),~i<j~should~be~twice~the~outcomes~of~(i,i).~This~is~well~explained~if~we\\ consider~all~36~cases.$